# Can Quanta change colour?

1. Jun 20, 2005

### cubed

I want a yes or no answer and a short explanation.

2. Jun 20, 2005

### cubed

hello, anyone there?

3. Jun 20, 2005

### cubed

just say yes or no....

4. Jun 20, 2005

### HallsofIvy

Why? What color are they normally?

5. Jun 20, 2005

### El Hombre Invisible

First, a caveat. A photon can be emitted, a photon can be absorbed. In between, you cannot measure the energy of the photon to see whether or not it can change - if you do, that's an absorption event and that same photon cannot be said to be observed again.

But other than that, no I don't think so. Different observers may well disagree on what the photon energy is, but a given photon of that energy will not change in transit to any one observer.

An observer at rest relative to a monochromatic light source will say that the source is emitting photons at a constant energy. A second moving with constant velocity away from the source will agree that it is emitting photons at a constant energy, but will disagree on what that energy is. A third observer accelerating away from the source will say the source is emitting photons of lower and lower energies. None are wrong, none are right. But this can't be explained by a photon changing colour in transit, as the photon would have to know which observer was going to observe it.

6. Jun 20, 2005

A very good answer, but what about Red-shifted photons? Now they DO change energy as they get stretched out. That's why the Microwave background radiation consists of microwaves, rather than the high energy photons that they originally were.

I posted a question on PF a year or so ago about where this energy went and there was some good discussion, but I never really got to grips with where actually it went. It is 'spread out over a larger spacetime' I suppose, but that is hard to put into context with the equation E=hf

7. Jun 20, 2005

### marlon

do you mean colour-charge or are you referring to colour as being the visible range of the EM energy spectrum. The first is changed by the strong force (respecting colour neutrality at all times) and the second is changed by Doppler-effect and friends :)

marlon

8. Jun 20, 2005

### dextercioby

The Compton effect shifts the frequency/energy of the photons,too.

Usually,the Doppler-FIZEAU (the "Doppler effect" is a name for the frequency shift of sound waves in the presence of Galilean boosts) is interpreted in terms of electromagnetic waves and not photons.Surely,one can discuss it for photons,but,back in ~1860,there were no such thing as photons,there was only LIGHT...

Daniel.

Daniel.

9. Jun 21, 2005

### El Hombre Invisible

Does a single photon get red-shifted? How can you say a photon measured as being at the red end was emitted nearer the blue end? Also, how does a possibly dimensionless particle get 'stretched'? I don't know - I'm asking (these aren't loaded questions). To my knowledge, photons from background radiation measured as microwaves would always have been measured as microwaves in that same reference frame. Of course, those frames would not have necessarily have been feasible back when the background radiation was bluer, but that's cosmology for you.

10. Jun 21, 2005

### RandallB

Chalk this one up to another Physics / QM paradox that just isn’t answered. From the Hubble expansion we can see how Red Shift occurs using wave theory but explaining it for an individual particle, a photon. How can it lose energy and change frequency as measured by observers in the same reference frame but separated by distance with expansion?

We can measure it. We can mathematically predict it. We can even describe it as “stretching” a partial. But that doesn’t explain it any better than we can explain entanglement or young’s double slit.

RB

11. Jun 22, 2005

It has been top of my list for a long time RandallB. Glad I'm not on my own!

12. Jun 22, 2005

### Hans de Vries

It doesn't. There's just a rule of thumb that galaxies at a larger distance have
a higher velocity relative to us.

$$f_{redshifted}\ \ =\ \ \sqrt{\frac{c-v}{c+v}}\cdot f$$

Regards, Hans

13. Jun 23, 2005

But the photons we receive DO have lower frequencies than when they set out.... otherwise we couldn't measure red shift! If the frequency is lower, the energy is lower..... How does your reply explain this???

14. Jun 23, 2005

### El Hombre Invisible

I think if you stick to what we can actually say for sure, you'd find that a hard statement to back up. You cannot measure the energy of a photon that is being emitted - only one being absorbed - i.e. by you and whatever you're detecting it with. That photon energy might have been measured differently by someone in a difference frame of reference (one in which the galaxy was getting closer or at rest, rather than receding). If you take the view that the photon you measure has the exact same energy as the photon emitted, IN THE SAME FRAME OF REFERENCE, it becomes less paradoxical. I think.

15. Jun 23, 2005

### RandallB

Red Shift is explained by E X P A N S I O N the word the same just gotten bigger because the space between the letters has gotten larger.
Expansion is important because we see Red Shifts that seem to show thinks traveling faster than light. But with expansion they don’t go FTL. It’s just over great distances and the extra time it takes to cover the expanded space makes it seem like the source was moving FTL.
With Expansion, (Hubble etc.) in hand and using WAVE theory, we can explain how Red Shift is occurring.

BUT when we look at the individual photon, which must lose energy to shift Red it no so clear how that happens. That's the Paradox.

Therefore I'm not 100% OK with idea of expansion itself.

Like I said - sure something to work on.
RB

16. Jun 23, 2005

### Hans de Vries

OK, but if it's a GR thing (cosmological redshift versus doppler redshift) then
you shouldn't talk about "In the same reference frame" The concept reference
frame gets lost in General Relativity. And, as far as I now, Energy is not
conserved in General Relativity at least not without an adaptation of the
definition of Energy.

John Baez has a lot of good stuf on GR on the web, for instance:

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

This link handles the cosmological redshift as well.

Regards, Hans.

17. Jun 23, 2005

I still think not. If you look at line spectra from distant Galaxies, all of the spectral lines are at lower frequencies than they should be. As these Spectral lines correspond to energy levels between atomic orbitals, either the atoms are lower in distant galaxies, OR the photon frequencies have changed.
As atoms should be the same everywhere (we have no evidence to contradict this) the photons MUST have been emitted at one frequency and absorbed at another. Of course, relativistically speaking, the photon was emitted and absorbed at the same instant (from the photon's frame of reference), but the energy level changes.

______________________________________________________________-

I posted this before seeing your last post Hans - sorry. I followed your suggested link and had a read. I'm sure the answer is there, but my Physics knowledge just isn't up to understanding it... Seems like I'll not be able to get a 'simplistic' answer to my query.

Last edited: Jun 23, 2005
18. Jun 24, 2005

### El Hombre Invisible

Two observers, one source of photons.

Observer A is a loooooong way from the source C, but remains at rest relative to it throughout the experiment. Observer B starts very near the source and moves towards A throughout the experiment. In the meantime, both are detecting photons from the source C.

A will observe no redshift - he knows that C is emitting photons with the same energy that he is observing them with. B, on the other hand, does observe redshift because he is moving away from the source of the photons. If B is accelerating towards A, he will see the redshift increase (the photon energies decrease) as he approaches A.

If you agree with this, you will see that it is the idea of photons 'changing colour' that is the paradox, as by the time B reaches A, both will be observing photons that have travelled the same path. If photon energies decreased between C and A, why would A not observe the lower energies? This is because in B's frame of reference, the photons being EMITTED are redder than they are in A's frame of reference.

Here the experiment is about the motion of bodies rather than the expansion of space in between them, but the same principal holds.

19. Jun 24, 2005

### ZapperZ

Staff Emeritus
Here's where the "problem" lies.

You are using the atomic energy levels calculated using non-relativistic QM. Implicitly, such calculations will work if you and the atom are in the same rest frame. However, you then use that same energy level and tried to do energy conservation in a different frame. Now we already know this will lead into a lot of problems even in classical mechanics (an object is stationary in one, and moving in the other, so where did the extra KE came from?)

What you should have done is to doppler shift the whole atom, and THEN, recalculate the apparent energy level in your frame. When you do this, the coulombic potential that you put into the Schrodinger equation will be shifted also (it is, after all, an EM property). You will find that the energy level is not the same good old level that you got in the rest frame.

Zz.

20. Jun 24, 2005

### RandallB

But Hans
It’s GR, cosmological red shift, Hubble expansion, that depend on "In the same reference frame" to keep from producing nonsense FTL events.
Let’s use El Hombre’s example:
This can only be true if we ignore Hubble for that “a loooooong way from the source C” comment. So let us insert the source of CBR here! Since we see a major red shift this source must not be in our reference frame, right. Just a little math should show us just how fast that frame is moving, SR should do. It the answer is, FTL!! And way Faster Then Light.
Now this will never do, how can we have FTL!
That’s where GR, Hubble, and Expansion sort’s it out. The point here is that the CRB Source and us as observer A are both in pretty much the same reference frame, therefore no FTL. How to account for the huge Red Shift with no FTL? With the magic of “expansion” between the source C to the observer A. Now frequency of the light has reduced considerable, undisputed and well observed. Easily understood using Wave thinking in the large Cosmos.

But now when El H and Adrian want to think about a individual photon to understand it all, things get a little uneasy. Sure, If you want use ZapperZ’s idea, where you need “to doppler shift the whole atom, and THEN, recalculate the apparent energy level” or in our case the photon.
Problem with that is applying a “doppler shift” to a individual photon is treating it like a wave, so although it might work in the math, for me it is far short of an explanation.
So what does this mean?
It means we have a PARADOX!